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3 years ago

Find the range and interquality for 5,9,6'8'7

Mathematics

2 answers:

ehidna [41]3 years ago

8 0

I don't know the interquality but the range is 4

Genrish500 [490]3 years ago

8 0

Range: 9 - 5 = 4

5, 6, 7, 8, 9

Median: 7

Lower Quartile: 5.5

Upper Quartile: 8.5

(IQR) Interquartile Range: 8.5 - 5.5 = 3

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2 years ago

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3 years ago

the area of a parallelogram shape land is on the square and length of its two adjacent sides are 25 m and 17 M find its diagonal

Alex787 [66]

Step-by-step explanation:

Draw diagonal AC

The triangle ABC has sides 17 and 25

Say AB is 17, BC is 25

Draw altitude on side BC from A , say h

h = 17 sin B

Area = 25*17 sin B = 408

sin B = 24/25

In ∆ ABC

Cos B = +- 7/25

= 625 + 289 — b^2 / 2*25*17

b^2 = 914 — 14*17 = 676

b = 26

h = 17*24/25 = 408/25 = 16.32

Draw the second diagonal BD

In ∆ BCD, draw altitude from D, say DE =h

BD^2 = h^2 + {(25 + sqrt (289 -h^2) }^2

BD^2 = 16.32^2 + (25 + 4.76)^2

= 885.6576 + 266.3424

BD = √ 1152 = 33.94 m

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2 years ago

Qué procedimiento debe realizar para convertir kilómetros a mililitros

natulia [17]

Answer:

multiplica el valor de volumen por 1e+15

Step-by-step explanation:

4 0

3 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers